Finding explicit form of the first characteristic class of a fiber bundle over $S^2$

These two problems are from an exam I've taken some time ago that I still didn't solve:

(1) Does there exist a fiber bundle over $S^2$ with fiber $S^1$, whose characteristic class is equal to

a) Twice a generator of the group $H^2(S^2,\pi_1(S^1))$,

b) Thrice a generator of the group $H^2(S^2,\pi_1(S^1))$.

(2) Construct the fiber bundle over $S^2$ with the fiber equal to $S^1$, whose first caracteristic class is equal to 17 times a generator of the group $H^2(S^2,\pi_1(S^1))$.

As I understand, bundles with the fiber $S^1$ is somehow related to bundles with the fiber $\mathbb{R}^2$. I also know that the first obstruction to creating a non-zero section of the tangent bundle over $S^2$ is twice the generator, but I don't fully understand this, nor understand how to piece it all together using the basic introduction to the obstruction theory I've been given.

Are there any tips on where to start? The main issue I feel is that I don't know how to explicitly write the first characteristic class as an element of the cohomology group.

• Hint : Let $V$ your vector bundle on $S^2$. You obtain a trivialization on the "north part" $S^2 \backslash \{S\}$ and the south part $S^2 \backslash \{N\}$, where $S,N$ are the South pole/North pole of $S^2$. Composing these gives you a map $S^2 \backslash \{N,S\} \to \text{Aut}(S^1)$. In fact, the isomorphism class of $V$ depends only on the homotopy class of this map ! So the characterstic class you wrote fully characterize the bundle, and moreover, since $S^2 \backslash \{N,S\}$ retracts on a circle, you are really looking at homotopy class of map $S^1 \to S^1$ i.e $\Bbb Z$... – Nicolas Hemelsoet Oct 8 '17 at 9:35
• ... so taking the map $z \mapsto z^n$ for $n \in \Bbb Z$ will gives you a fiber bundle, with first characteristic class equal to $n$ times the generator of $H^2(S^2, \Bbb Z)$. – Nicolas Hemelsoet Oct 8 '17 at 9:35
• In a similar manner, note that the Hopf bundle $\eta:S^3\rightarrow S^2$ represents the $S^1$-bundle with first Chern class a generator of $H^2(S^2)$. Now form the associated bundles $S^3\times_{S^1(n)}S^1=(S^3\times S^1)/[(z\lambda,\rho)\sim (z,\lambda^{n}\rho)]$ using the Borel construction. It's not difficult to show that this is principal ($S^1$ is abelian), and its not difficult to calculate its transition function, and hence its Chern class. – Tyrone Oct 8 '17 at 11:48
• @NicolasHemelsoet I'm not sure I understand how the composition leads to the map you described. If we denote the trivializations on the north and south parts as $f_N$ and $f_S$, then $f_Nf_S^{-1}:\mathbb{R}^2\times S^1 \rightarrow \mathbb{R}^2\times S^1$ induces a homeomorphism $S^1\rightarrow S^1$ by restriction. Here I denoted $S^2-\{N\}$ as $\mathbb{R}^2$. So if you initially restrict $f_S^{-1}$ to just one point, is that how you then associate an element of $Aut(S^1)$ to the map? (am I correct in understanding $Aut(S^1)$ to be the group of homeomorphisms of $S^1$ preserving orientations?). – Maksim Dolgikh Oct 8 '17 at 15:47
• Yes, but $f_N$ and $f_S$ has commun domain $\Bbb R^2 \backslash \{0\}$ and not $\Bbb R^2$, i.e $S^2$ minus two point. So $f_Nf_S^{-1} : \Bbb R^2 \backslash \{0\} \times S^1 \to \Bbb R^2 \backslash \{0\} \times S^1$. Moreover, this map has to be a bundle map, i.e it is the identify on $\Bbb R^2 \backslash \{0\}$. So really, your map is on the form $(x,\theta) \mapsto (x, f(x)(\theta))$ where $f(x) \in Aut(S^1)$. Moreover, as $\Bbb R^2 \backslash \{0\}$ retracts on $S^1$ we get indeed a map $S^1 \to Aut(S^1)$. I hope it is clear, you can google "clutching functions", I believe the book ... – Nicolas Hemelsoet Oct 8 '17 at 16:00